Circuits Current Resistance & Ohm’s Law Resistors in Series, in Parallel, in addition to in

Circuits Current Resistance & Ohm’s Law Resistors in Series, in Parallel, in addition to in www.phwiki.com

Circuits Current Resistance & Ohm’s Law Resistors in Series, in Parallel, in addition to in

Sharpe, Don, Founder & Publisher has reference to this Academic Journal, PHwiki organized this Journal Circuits Current Resistance & Ohm’s Law Resistors in Series, in Parallel, in addition to in combination Capacitors in Series in addition to Parallel Voltmeters & Ammeters Resistivity Power & Power Lines Fuses & Breakers Bulbs in Series & Parallel Electricity The term electricity can be used to refer to any of the properties that particles, like protons in addition to electrons, have as a result of their charge. Typically, though, electricity refers to electrical current as a source of power. Whenever valence electrons move in a wire, current flows, by definition, in the opposite direction. As the electrons move, their electric potential energy can be converted to other as long as ms like light, heat, in addition to sound. The source of this energy can be a battery, generator, solar cell, or power plant. Current By definition, current is the rate of flow of positive charge. Mathematically, current is given by: If 15 C of charge flow past some point in a circuit over a period of 3 s, then the current at that point is 5 C/s. A coulomb per second is also called an ampere in addition to its symbol is A. So, the current is 5 A. We might say, “There is a 5 amp current in this wire.” It is current that can kill a someone who is electrocuted. A sign reading “Beware, High Voltage!” is really a warning that there is a potential difference high enough to produce a deadly current.

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Charge Carriers & Current A charge carrier is any charged particle capable of moving. They are usually ions or subatomic particles. A stream of protons, as long as example, heading toward Earth from the sun (in the solar wind) is a current in addition to the protons are the charge carriers. In this case the current is in the direction of motion of protons, since protons are positively charged. In a wire on Earth, the charge carriers are electrons, in addition to the current is in the opposite direction of the electrons. Negative charge moving to the left is equivalent to positive charge moving to the right. The size of the current depends on how much charge each carrier possesses, how quickly the carriers are moving, in addition to the number of carriers passing by per unit time. wire electrons I protons I A Simple Circuit A circuit is a path through which an electricity can flow. It often consists of a wire made of a highly conductive metal like copper. The circuit shown consists of a battery ( ), a resistor ( ), in addition to lengths of wire ( ). The battery is the source of energy as long as the circuit. The potential difference across the battery is V. Valence electrons have a clockwise motion, opposite the direction of the current, I. The resistor is a circuit component that dissipates the energy that the charges acquired from the battery, usually as heat. (A light bulb, as long as example, would act as a resistor.) The greater the resistance, R, of the resistor, the more it restricts the flow of current. Building Analogy To underst in addition to circuits, circuit components, current, energy trans as long as mations within a circuit, in addition to devices used to make measurements in circuits, we will make an analogy to a building. Continued V R I

Building Analogy Correspondences Battery Elevator that only goes up in addition to all the way to the top floor Voltage of battery Height of building Positive charge carriers People who move through the building en masse (as a large group) Current Traffic (number of people per unit time moving past some point in the building) Wire w/ no internal resistance Hallway (with no slope) Wire w/ internal resistance Hallway sloping downward slightly Resistor Stairway, ladder, fire pole, slide, etc. that only goes down Voltage drop across resistor Length of stairway Resistance of resistor Narrowness of stairway Ammeter Turnstile (measures traffic without slowing it down) Voltmeter Tape measure ( as long as measuring changes in height) Current in addition to the Building Analogy In our analogy people correspond to positive charge carriers in addition to a hallway corresponds to a wire. So, when a large group of people move together down a hallway, this is like charge carriers flowing through a wire. Traffic is the rate at which people are passing, say, a water fountain in the hall. Current is rate at which positive charge flows past some point in a wire. This is why traffic corresponds to current. Suppose you count 30 people passing by the fountain over a 5 s interval. The traffic rate is 6 people per second. This rate does not tell us how fast the people are moving. We don’t know if the hall is crowded with slowly moving people or if the hall is relatively empty but the people are running. We only know how many go by per second. Similarly, in a circuit, a 6 A current could be due to many slow moving charges or fewer charges moving more quickly. The only thing as long as certain is that 6 coulombs of charge are passing by each second. Battery & Resistors in addition to the Building Analogy elevator top floor hallway: high Ugrav V R bottom floor hallway: zero Ugrav staircase flow of + charges + – flow of people Our up-only elevator will only take people to the top floor, where they have maximum potential in addition to , thus, where they are at the maximum gravitational potential. The elevator “energizes” people, giving them potential energy. Likewise, a battery energizes positive charges. Think of a 10 V battery as an elevator that goes up 10 stories. The greater the voltage, the greater the difference in potential, in addition to the higher the building. As reference points, let’s choose the negative terminal of the battery to be at zero electric potential in addition to the ground floor to be at zero gravitational potential. Continued

Battery & Resistors in addition to the Building (cont.) Current flows from the positive terminal of the battery, where + charges are at high potential, through the resistor where they give up their energy as heat, to the negative terminal of the battery, where they have zero potential energy. The battery then “lifts them back up” to a higher potential. The charges lose no energy moving the a length of wire (with no internal resistance). Similarly, people walk from the top floor where they are at a high potential, down the stairs, where their potential energy is converted to waste heat, to the bottom floor, where they have zero potential energy. The elevator them lifts them back up to a higher potential. The people lose no energy traveling down a (level) hallway. elevator top floor hallway: high Ugrav V R bottom floor hallway: zero Ugrav staircase flow of + charges + – flow of people Resistance Resistance is a measure of a resistors ability to resist the flow of current in a circuit. As a simplistic analogy, think of a battery as a water pump; it’s voltage is the strength of the pump. A pipe with flowing water is like a wire with flowing current, in addition to a partial clog in the pipe is like a resistor in the circuit. The more clogged the pipe is, the more resistance it puts up to the flow of water trying to flow through it, in addition to the smaller that flow will be. Similarly, if a resistor has a high resistance, the current flowing it will be small. Resistance is defined mathematically by the equation: V = I R Resistance is the ratio of voltage to current. The current flowing through a resistor depends on the voltage drop across it in addition to the resistance of the resistor. The SI unit as long as resistance is the ohm, in addition to its symbol is capital omega: . An ohm is a volt per ampere: 1 = 1 V / A The Voltage Lab (scroll down) Resistance in addition to Building Analogy In our building analogy we’re dealing with people instead of water molecules in addition to staircases instead of clogs. A wide staircase allows many people to travel down it simultaneously, but a narrow staircase restricts the flow of people in addition to reduces traffic. So, a resistor with low resistance is like a wide stairway, allowing a large current though it, in addition to a resistor with high resistance is like a narrow stairway, allowing a smaller current. V = 12 V R = 6 I = 2 A V = 12 V R = 3 I = 4 A Narrow staircase means reduced traffic. Wide staircase means more traffic.

Ohm’s Law The definition of resistance, V = I R, is often confused with Ohm’s law, which only states that the R in this as long as mula is a constant. In other words, the resistance of a resistor is a constant no matter how much current is flowing through it. This is like saying a clog resists the flow of water to the same extent regardless of how much water is flowing through it. It is also like saying a the width of a staircase does not change: no matter what rate Georg Simon Ohm 1789-1854 people are going downstairs, the stairs hinder their progress to the same extent. In real life, Ohm’s law is not exactly true. It is approximately true as long as voltage drops that aren’t too high. When voltage drops are high, so is the current, in addition to high current causes more heat to generated. More heat means more r in addition to om thermal motion of the atoms in the resistor. This, in turn, makes it harder as long as current to flow, so resistance goes up. In the circuit problems we do we will assume that Ohm’s law does hold true. Ohmic vs. Nonohmic Resistors If Ohm’s law were always true, then as V across a resistor increases, so would I through it, in addition to their ratio, R (the slope of the graph) would remain constant. In actuality, Ohm’s law holds only as long as currents that aren’t too large. When the current is small, not much heat is produced in a real, so resistance is constant in addition to Ohm’s law holds (linear portion of graph). But large currents cause R to increase (concave up part of graph). ohmic non-ohmic I V Ohmic Resistor Real Resistor I V Series & Parallel Circuits Resistors in Series I V R1 R2 R3 V I R1 R2 R3 When several circuit components are arranged in a circuit, they can be done so in series, parallel, or a combination of the two. Resistors in Parallel

Resistors in Series: Building Analogy To go from the top to the bottom floor, all people must take the same path. So, by definition, the staircases are in series. With each flight people lose some of the potential energy given to them by the elevator, expending all of it by the time they reach the ground floor. So the sum of the V drops across the resistors the voltage of the battery. People lose more potential energy going down longer flights of stairs, so from V = I R, long stairways correspond to high resistance resistors. The double waterfall is like a pair of resistors in series because there is only one route as long as the water to take. The longer the fall, the greater the resistance. 3 steps 6 steps 11 steps Elevator (battery) R1 R2 R3 R1 R2 Equivalent Resistance in Series I V R1 R2 R3 I V Req If you were to remove all the resistors from a circuit in addition to replace them with a single resistor, what resistance should this replacement have in order to produce the same current This resistance is called the equivalent resistance, Req. In series Req is simply the sum of the resistances of all the resistors, no matter how many there are: Req = R1 + R2 + R3 + · · · Mnemonic: Resistors in Series are Really Simple. Proof of Series Formula V1 + V2 + V3 = V (energy losses sum to energy gained by battery) V1= I R1, V2= I R2, in addition to V3= I R3 ( I is a constant in series) I V R1 R2 R3 } V1 } V2 } V3 I R1 + I R2 + I R3 = I Req ( substitution) R1 + R2 + R3 = Req ( divide through by I ) I V Req

4 1. Find Req 2. Find Itotal 3. Find the V drops across each resistor. 12 0.5 A 2 V, 1 V, in addition to 3 V (in order clockwise from top) Solution on next slide Series Sample 4 1. Since the resistors are in series, simply add the three resistances to find Req: Req = 4 + 2 + 6 = 12 2. To find Itotal (the current through the battery), use V = I R: 6 = 12 I. So, I = 6/12 = 0.5 A Series Solution 3. Since the current throughout a series circuit is constant, use V = I R with each resistor individually to find the V drop across each. Listed clockwise from top: V1 = (0.5)(4) = 2 V V2 = (0.5)(2) = 1 V V3 = (0.5)(6) = 3 V Note the voltage drops sum to 6 V. 9 V 1 7 6 3 3. Find the V drop across each resistor. 17 0.529 A V1 = 3.2 V V2 = 0.5 V V3 = 3.7 V V4 = 1.6 V check: V drops sum to 9 V. Series Practice 1. Find Req 2. Find Itotal

Resistors in Parallel: Building Analogy Suppose there are two stairways to get from the top floor all the way to the bottom. By definition, then, the staircases are in parallel. People will lose the same amount of potential energy taking either, in addition to that energy is equal to the energy the acquired from the elevator. So the V drop across each resistor equals that of the battery. Since there are two paths, the sum of the currents in each resistor equals the current through the battery. A wider staircase will accommodate more traffic, so from V = I R, a wide staircase corresponds to a resistor with low resistance. The double waterfall is like a pair of resistors in parallel because there are two routes as long as the water to take. The wider the fall, the greater the flow of water, in addition to lower the resistance. R1 R2 Elevator (battery) Equivalent Resistance in Parallel V I R1 R2 R3 I1 I2 I3 I1 + I2 + I3 = I (currents in branches sum to current through battery ) V = I1 R1, V = I2 R2, in addition to V = I3 R3 (V is a constant in parallel) V R1 V R2 + V R3 V Req + = (substitution) 1 R1 R2 + R3 Req + = (divide through by V ) 1 1 1 I V Req This as long as mula extends to any number of resistors in parallel. 4 6 15 V 3. Find the current through, in addition to voltage drop across, each resistor. 2.4 6.25 A It’s a 15 V drop across each. Current in middle branch is 3.75 A; current in right branch is 2.5 A. Note that currents sum to the current through the battery. Parallel Example 1. Find Req 2. Find Itotal Solution on next slide

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4 6 1. 1/Req= 1/R1 + 1/R2 = 1/4 + 1/6 = 6/24 + 4/24 = 5/12 Req = 12/5 = 2.4 2. Itotal = V / Req = 15 / (12/5) = 75/12 = 6.25 A 15 V 3. The voltage drop across each resistor is the same in parallel. Each drop is 15 V. The current through the 4 resistor is (15 V)/(4 ) = 3.75 A. The current through the 6 resistor is (15 V)/(6 ) = 2.5 A. Check: Parallel Solution Itotal I1 I2 Itotal = I1 + I2 12 16 8 24 V I1 = 2 A I2 = 1.5 A I3 = 3 A V drop as long as each is 24 V. 13/2 A 48/13 = 3.69 Parallel Practice 3. Find the current through, in addition to voltage drop across, each resistor. 1. Find Req 2. Find Itotal Hint: First find the V drop over the 4 resistor next to the battery. This resistor is in series with the rest of the circuit. Subtract this V drop from that of the battery to find the remaining drop along any path. 8.5 1.0588 A Combo Sample 1. Find Req 2. Find Itotal 3. Find the current through, in addition to voltage drop across, the highlighted 9 V resistor. Solutions 0.265 A, 2.38 V

We simplify the circuit a section at a time using the series in addition to parallel as long as mulae in addition to use V = I R in addition to the end. The units have been left off as long as clairy. Combo Solution: Req & Itotal 9 V Req = 8.5 I total =1.0588 A To find the current in the red resistor we must find the voltage drop across its branch. Working from the simplified circuit on the last slide, we see that the resistor next to the battery is in series with the rest of the circuit, which is a 4.5 equivalent. The total current flows through the 4 , so the V drop across it is 1.0588(4) = 4.235 V. Subtracting from 9 V, this leaves 4.765 V across the 4.5 equivalent. There is the same drop across each parallel branch within the equivalent. We’re interested in the left branch, which has 18 of resistance in it. This means the current through the left branch is 4.765 / 18 = 0.265 A. This is Combo Solution: V Drops & Current the current through the red resistor. The voltage drop across it is 0.265(9) = 2.38 V. Note that this is half the drop across the left branch. This must be the case since 9 is half the resistance of this branch. 3. Find the current through, in addition to voltage drop across, the resistor R. 1. Find Req 2. Find Itotal Combo Practice Each resistor is 5 , in addition to the battery is 10 V. R 6.111 1.636 A 0.36 A

Light Bulbs in Parallel Light bulbs are intended in addition to labeled as long as parallel circuits, since that’s how are homes are wired. Suppose we hook up 3 bulbs of different wattages in parallel as shown. The filament of each bulb acts as a resistor. Each bulb has same potential difference across it, but the currents going the each must be different. Otherwise, they would be equally bright. As you would expect, the 100 W bulb is the brightest. From P = I V, the 100 W bulb must have the highest current going through it (since V is constant). From V = I R, the 100 W bulb must have the filament with the lowest resistance. Note that if one bulb is removed, the others still shine. In summary, in parallel: V I R60 R75 R100 I60 I75 I100 60 W 75 W 100 W I60 < I75 < I100 R100 < R75 < R60 V = constant Light Bulbs in Series I V R60 R75 R100 60 W 75 W 100 W Let’s place the same 3 bulbs in series now. From P = I 2 R, the power output of any bulb is proportional to its resistance (since each has the same current flowing through it). On the last slide we concluded that bulbs labeled with higher wattages have lower resistances. The resistances of their filaments remain the same no matter how they are wired. This means the 100 W bulb will be the dimmest, in addition to the 60 W bulb will be the brightest. Note that if any bulb is P100 < P75 < P60 R100 < R75 < R60 I = constant high R, bright low R, dim removed now, all bulbs go out. Also note that the power consumption stamped on a bulb is only correct if the bulb is connected in parallel with at a certain voltage. In summary, in series: CREDITS Ohm picture: http://hubcap.clemson.edu/~asommer/ohm.html Voltage Lab: http://jersey.voregon.edu.edu/vlab/Voltage/ Color code picture: http://webhome.idirect.com/~jadams/electronic/resist-codes.html Color Code Link: http://www.electrician.com/resist-calc/resist-calc.htm Ohm picture: http://hubcap.clemson.edu/~asommer/ohm.html Voltage Lab: http://jersey.voregon.edu.edu/vlab/Voltage/ Color code picture: http://webhome.idirect.com/~jadams/electronic/resist-codes.html Color Code Link: http://www.electrician.com/resist-calc/resist-calc.htm

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